Tratar servlet.error no controller [RESOLVIDO]
19/08/2009 00:00
"500"(controller:'error',action:'getError')
<strong>Error ${request.'javax.servlet.error.status_code'}:</strong> ${request.'javax.servlet.error.message'.encodeAsHTML()}<br/>
<strong>URI:</strong> ${request.'javax.servlet.error.request_uri'}<br/>
<strong>Exception Message:</strong> ${exception.message?.encodeAsHTML()} <br />
import javax.servlet.http.HttpServletRequest
import javax.servlet.http.HttpServlet
class ErrorController extends HttpServlet{
def getError(HttpServletRequest req){
String code = ""
Object codeObj = null
codeObj = req.getAttribute("javax.servlet.error.status_code")
code = codeObj.toString()
flash.error = "${code}"
redirect(controller:'aluno',action:'list')
}
}
class ErrorController {
def getError = {
String code = "${request.'javax.servlet.error.status_code'}"
String causa = "${request.'javax.servlet.error.message'.encodeAsHTML()}"
String uri = "${request.'javax.servlet.error.request_uri'}"
String erro = "Erro: ${code} causado por: ${causa} em: ${uri}"
flash.error = "${erro}"
redirect(controller:'aluno',action:'list')
}
}
causa = "${request.'javax.servlet.error.message'.encodeAsHTML()}"
uri = "${request.'javax.servlet.error.request_uri'}"
causa = request.'exception'
uri = request.'javax.servlet.forward.request_uri'
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